/* 

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

 

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1
示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3
 

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

*/

/**
 * @param {character[][]} grid
 * @return {number}
 */

// “沉没”法
var numIslands = function (grid) {
    let count = 0;

    // dfs 深度优先遍历
    let sink = function (row, col) {
        if (row < 0 || row >= grid.length || col < 0 || col >= grid[row].length || grid[row][col] === "0") {
            return;
        }

        grid[row][col] = "0";

        sink(row + 1, col);
        sink(row - 1, col);
        sink(row, col + 1);
        sink(row, col - 1);
    }

    for (let row = 0; row < grid.length; row++) {
        for (let col = 0; col < grid[row].length; col++) {
            if (grid[row][col] === "1") {
                count++;
                sink(row, col);
            }
        }
    }

    return count;
};